Answer
(a) $A(t)=5000e^{0.09t}$
(b) See the graph below.
(c) At the end of $17^{th}$ and the beginning of $18^{th}$ years. It will pass the amount in $18^{th}$ year.
Work Step by Step
(a) $A(t)=Pe^{rt}=5000e^{0.09t}$
(b) We can sketch the graph using the function calculated in (a) and graphing calculator. See the image above.
(c) As we can see on the graph above the investment reaches $\$25,000$ at the end of $17^{th}$ and the beginning of $18^{th}$ years.
